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- Question 9 Exercise 6.3
- actly contain four men and four women are: \begin{align}{ }^7 C_4 \cdot{ }^6 C_4&=\dfrac{7 !}{(7-4) ! 4 !} \cdot \dfrac{6 !}{(6-4)}\\\ &= 525\end{align} =====Question 9(ii)===== An $8$-persons committ... s case the total number of committees are: \begin{align}{ }^7 C_2 \cdot{ }^6 C_6&=\dfrac{7 !}{(7-2) ! 2 !} \cdot \dfrac{6 !}{(6-6) ! 6 !}\\ &=21\end{align} If committee contains $3$ men then it will cont
- Question 13 Exercise 6.2
- are $L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutations are} &=\left(... ac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\ &=37,800 \end{align} Begin with $\mathrm{E}$ If we have to pick the ... $ are $L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{Number of permulations are} &=\left(\begin{... rac{9 !}{3 ! \cdot 2 ! \cdot 2 !}\\ &=15,120 \end{align} =====Question 13(ii)===== Find the number of pe
- Question 7 Exercise 6.4
- The sample space rolling a pair of dice is \begin{align}S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin... 3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \t... imes 6=36$$ doublet of even numbers. Let \begin{align}A&=\{(2,2),(4,4),(6,6)\}\\ n(A)&=3\end{align} Hence the possibility of getting doublet of an even number
- Question 4 Exercise 6.4
- the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ \text{then} n(S)&=2^3=8\end{align} When all heads. Let $$A=\{H H H\}$$ then $$n(A)... the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ \text{then} n(S)&=2^3=8\end{align} When two heads Let $$B=\{H T$ T.THT.TTH $\}$$ t
- Question 9 Exercise 6.5
- at both will be selected. ====Solution==== \begin{align} P(\text { Ajmal scicction })&=\dfrac{1}{7} \\ \R... (\text { Bushra not selected })&=\dfrac{4}{5}\end{align} Both are selected Since the selection of one do... ore these two events are independent. Thus \begin{align}P(\text { Both are selected })&=P(A) \times P(B) ... dfrac{1}{7} \times \dfrac{1}{5}=\dfrac{1}{35}\end{align} =====Question 9(ii)===== A'jmal and Bushra appe
- Question 1 and 2 Exercise 6.1
- 0 !}{3 ! .3 ! \cdot 4 !}$ ====Solution==== \begin{align}\dfrac{10 !}{3 ! \cdot 3 ! \cdot 4 !}&=\dfrac{10.... }\\ &=\dfrac{10.9 .8 .7 .5}{3.2 .1}\\ &=4200 \end{align} =====Question 1(ii)===== Evaluate the $\dfrac{3 !+4 !}{5 !-4 !}$ ====Solution==== \begin{align}\dfrac{3 !+4 !}{5 !-4 !}&=\dfrac{3 !+4.3 !}{5.4 !... ac{3 !(5)}{4 \cdot 3 !(4)}\\ &=\dfrac{5}{16} \end{align} =====Question 1(iii)===== Evaluate the $\dfrac{
- Question 1 and 2 Exercise 6.5
- === We know by addition law of probability \begin{align} P(A \cup B)&=P(A)+P(B)-P(A \cap B) \\ \Rightarrow P(A \cap B)&=P(A)+P(B)-P(A \cup B) \end{align} Substituting $P(A), P(B)$ and $P(A \cup B)$, we ... ntary events $$P(B)=1-P(\bar{B})$$ Putting \begin{align}P(\bar{B})&=\dfrac{5}{8}\\ P(B)&=1-\dfrac{5}{8}=\dfrac{3}{8}\end{align} Also $$P(\bar{A})=1-P(A)$$ Putting $P(A)=\dfrac
- Question 2 Exercise 6.3
- ^n C_r=35$. ====Solution==== We are given: \begin{align} &^n P_r=\dfrac{n !}{(n-r) !}=840 ....(i)\\ &^n C_r=\dfrac{n !}{(n-r) ! r !}=35....(ii)\end{align} Dividing Eq.(i) by Eq.(ii) \begin{align}\dfrac{n !}{(n-r) !} \cdot \dfrac{(n-r) ! r !}{n !}&=\dfrac{840}{35}\\ r!&=24\\ \text{or}\quad r &=4\end{align} Putting $r=4$ in Eq.(ii), we get \begin{align} &
- Question 1 and 2 Exercise 6.2
- i)===== Evaluate $^6 P_6$ ====Solution==== \begin{align}^6 P_6&=\dfrac{6 !}{(6-6) !}\\ &=6 !=720\end{align} =====Question 1(ii)===== Evaluate $^{20} P_2$ ====Solution==== \begin{align}^{20} P_2&=\dfrac{20 !}{(20-2) !}\\ &=\dfrac{20.19 .18 !}{18 !}\\ &=20 \times 19=380\end{align} =====Question 1(iii)===== Evaluate $^{16} P_3$ =
- Question 3 and 4 Exercise 6.5
- ive, therefore $A \cap B=\emptyset$. Thus \begin{align}P(A \cup B)&=P(A)+P(B)\\ \Rightarrow P(B)&=P(A \cup B)-P(A)\\ &=0.6-.0 .5=0.1 \end{align} =====Question 4===== A bag contains $30$ ticket... n==== Total numbers written on tickets are \begin{align}S&=\{1,2,3, \ldots, 50\} \text { so }\\ n(S)&=50 \end{align} Let \begin{align}A \{odd \,numbers \}&=\{1,3,5,.
- Question 5 Exercise 6.1
- We are taking L.H.S of the above equation \begin{align}\dfrac{(2 n) !}{n !}&=\dfrac{1}{n !}[(2 n)(2 n-1)... )\\ &(2 n-(2 n-3))(2 n-(2 n-2))(2 n-(2 n-1))]\end{align} In the L.H.S of the above equation are total $2 n$ terms \begin{align}\dfrac{(2 n) !}{n !}&=\dfrac{1}{n !}[(2 n)(2 n-1)... 4.2] {[(2 n-1)(2 n-3)(2 n-5) \ldots 5.3 .1]} \end{align} Each underlined brackets contain $n$ terms, that
- Question 1 Exercise 6.4
- a $5$ ? ====Solution==== Rolling a $5$ Let \begin{align}A&=\{5\}\\ P(A)&=\dfrac{n(A)}{n(S)}\\ &=\dfrac{1}{6} \end{align} =====Question 1(b)===== Let $S=\{1,2,3,4,5,6\}$... ion==== Rolling a number less than $1$ Let \begin{align}B&=\{\}\\ &=\phi \text{then}\\ P(B)&=\dfrac{n(B)}{n(S)}\\ &=\dfrac{0}{6}\\ &=0\end{align} =====Question 1(c)===== Let $S=\{1,2,3,4,5,6\}$
- Question 3 & 4 Exercise 6.1
- ing the L.H.S of the above given equation. \begin{align}\dfrac{1}{6 !}+\dfrac{2}{7 !}+\dfrac{3}{8 !}&=\df... \ & =\dfrac{56+16+3}{8 !}\\ &=\dfrac{75}{8 !}\end{align} =====Question 3(ii)===== Prove that $\dfrac{(n+... ing the L.H.S of the above given equation. \begin{align}\dfrac{(n+5) !}{(n+3) !}&=\dfrac{(n+5)(n+4)(n+3) !}{(n+3) !} \\ & =(n+5)(n+4)\\ &=n^2+9 n+20\end{align} =====Question 4(i)===== Find the value of $n$,
- Question 10 Exercise 6.2
- ays these five students can be seated are: \begin{align}^8 P_5&=\dfrac{8 !}{(8-5) !}\\ &=\dfrac{8 \cdot 7... ot 6 \cdot 5 \cdot 4 \cdot 3 !}{3 !}\\ &=6720\end{align} If certain two students insist to sit next to ea... In this case the total number of ways are: \begin{align}^2 P_2 \times^7 P_4&=2 \times \dfrac{7 !}{(7-4) !... =2 \times\dfrac{7.6 .5 .4 .3 !}{3 !}\\ &=1680\end{align} =====Question 10(ii)===== In how many ways can
- Question 12 Exercise 6.2
- f different words using all at a time are: \begin{align} \left(\begin{array}{c} n \\ m 1 \end{array}\rig... 6 \cdot 5 \cdot 4 \cdot 3 !}{3 !}\\ &=6,720 \end{align} =====Question 12(ii)===== How many different wo... s the total number of different words are: \begin{align} \left(\begin{array}{c} n \\ m_1, m_2, m_3 \end{a... c{10 !}{2 ! \cdot 3 ! \cdot 2 !}\\ &=151,200 \end{align} =====Question 12(iii)===== How many different w